Question

QUESTION 5

1. CaCO3 (s) ⇌ CaO (s) + CO2 (g)

   At 599 °C, the reaction reaches equilibrium.

   If P_CO2 =3.01 atm, what is Kc value ?

   *Please report 3 significant figures. Numbers only, No unit. No scientific notation.

Answer 1

Question.5.Ans :-

Given reacion is :

CaCO3 (s) <--------------------> CaO (s) + CO2 (g)

Expression of Pressure Equilibrium constant i.e. Kp (which is equal to the product of the partial pressure of products divided by product of the partial pressure of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).

Kp = PCO2 (g)

Given PCO2 (g) = 3.01 atm

Therefore,

Kp = 3.01 atm

Now, the relationship between Kp and Kc (Concentration equilibrium constant) is :

Kp = Kc(RT)Δng

Kc = Kp(RT)-Δng

Where,

Δng = Change in number of moles of gases in the reaction = 1 - 0 = 1

R = Gas constant = 0.0821 L atm K-1mol-1

T = Temperature = 273 + 599 = 872 K

Kc = (3.01).(0.0821 L atm K-1mol-1 x 872 K)-1

Kc = (3.01) / (0.0821 L atm K-1mol-1 x 872 K)

Kc = (3.01) / (0.0821 L atm K-1mol-1 x 872 K)

Kc = 0.0420

Therefore, value of Kc = 0.0420